A History of the Arf-Kervaire Invariant Problem

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A History of the Arf-Kervaire Invariant Problem A History of the Arf-Kervaire Invariant Problem Victor P. Snaith “I know what you’re thinking about,” said or dyadic formula for n. For example, 17 = 1 + 24, Tweedledum; “but it isn’t so, nohow.” “Con- 35 = 1 + 2 + 25, and 60 = 22 + 23 + 24 + 25. The trariwise,” continued Tweedledee, “if it was binary formula may be depicted as a string of 0’s so, it might be; and if it were so, it would be; and 1’s that records in the i-th place whether or but as it isn’t; it ain’t. That’s logic.” not 2i appears in the binary formula for n. Here is a sample table of binary strings: —Through the Looking Glass, by Lewis Carroll (aka Charles Lutwidge Dodgson) [8] 17 10001 35 110001 Typing “Invariant theory” into Wikipedia yields the 60 001111 theory of functions like x1x2 +x1x3 +x2x3 which are unaltered by permuting the variables. In algebraic If we express each of n1; : : : ; nk as a row in a topology, particularly post-1950, a different notion similar table of binary strings, we define the Nim of “invariant” emerged. This use of invariant (e.g., invariant to be the string of integers given by the Hopf invariant, Arf-Kervaire invariant, λ-invariant) column sums. The Nim invariant of 17; 35; 60 is denotes an algebraic quantity that gives a partial (2; 1; 1; 1; 2; 2). If every entry in the Nim invariant answer to a topological question. is even (which we shall call the Nim condition) after Often invariants in this sense are very technical, a player’s turn, then the answer to the question is both in their context and in their construction. yes; otherwise it is no. However, a very simple invariant occurs in the game If the Nim condition holds after Player A’s turn, of Nim ([31], pp. 36–38). In the 1960s this game then either A has won or any move by Player B was popular among students due to its enigmatic destroys the Nim condition. In particular, Player B appearance in Alain Resnais’s 1961 avant-garde has not won! Conversely, if the Nim condition movie L’Année Dernière à Marienbad. does not hold after Player B’s turn, then Player A A set of matchsticks is divided arbitrarily into can restore it by the following algorithm: Player A several heaps. Two players play alternately. A play inspects the string of column sums from right to consists of selecting a heap and removing from it left to find the first odd column sum. Suppose this any (nonzero) number of matchsticks. The winner is the column corresponding to 2i. Then Player A is the player whose move leaves no remaining chooses a row in which there is a 1 in the i-th matchsticks. The question which the Nim invariant column. Player A takes some matchsticks from the answers is, If my opponent and I play out of our pile corresponding to this row. There is always a skins, will I win? number of matchsticks which Player A can remove Suppose there are k heaps of matchsticks of from this pile in order to restore the Nim condition. sizes n1; : : : ; nk. Recall that every positive integer n For example, in the table above, the 23-column has can be written in one and only one way as the sum an odd sum, and Player A finds a 1 in the bottom of distinct powers of 2. This is called the binary row. Subtracting 23 + 2 = 10 from 60 changes the table to: Victor P. Snaith is emeritus professor of mathematics at the University of Sheffield. His email address is v.snaith@ 17 10001 sheffield.ac.uk. 35 110001 DOI: http://dx.doi.org/10.1090/noti1030 50 010011 1040 Notices of the AMS Volume 60, Number 8 whose column sum is (2; 2; 0; 0; 2; 2). With this The Nim and Arf invariants fortuitously give a algorithm Player A has restored the Nim condition necessary and sufficient answer to their mathe- and reduced the total number of matchsticks. Since matical questions. As the mathematical subtlety Player B cannot win, playing perfectly, Player A of the question deepens, one must often settle must do so. for invariants which give only partial information. The Arf invariant, which occurred first in algebra Algebraic topology is littered with examples of the [5], requires a little more mathematical background. calculational device known as a spectral sequence. Let V be the n-dimensional vector space over the A spectral sequence is an invariant (some poetic field F2 of two elements. In more concrete terms, license may be needed here) whose output can let V be the set of n-tuples x = (x1; x2; : : : ; xn) often be highly ambiguous. Spectral sequences, in which each xi is equal to 0 or 1. Recall that like Marmite, are either loved or hated, and in the “addition” on V is defined by setting x + y equal to 1960s, graduate courses about them were regularly the n-tuple whose j-th entry equals 0 if xj + yj is inflicted on unwilling students in a variety of re- even and equals 1 otherwise. F2 is the example in search areas. One of the most famous is the Adams which V consists of 1-tuples. spectral sequence, invented in the 1950s by Frank ! A quadratic form is a function q : V - F2 such Adams [1] (unfortunately universally abbreviated that q((0;:::; 0)) = 0 and the associated function to ASS), which turns cohomological algebra into Q : V ×V -! F given by Q(x; y) = q(x+y)−q(x)− 2 calculational information about stable homotopy q(y) satisfies Q(x; y + z) = Q(x; y) + Q(x; z). This groups. Typically a spectral sequence comprises an function of two V -variables is called an F -bilinear 2 infinite family of abelian groups, calculable by an form. It is symmetric and symplectic, which means algebraic algorithm, together with maps between Q(x; y) = Q(y; x) and Q(x; x) = 0 for all x; y in V. them called differentials. The ambiguity arises The function q is called a quadratic refinement from the fact that. even if one knew the identity of Q. Three examples when n = 2 are given by 0 2 2 of all the differentials, the algorithm for their use q(x1; x2) = x1x2, q (x1; x2) = x1 + x1x2 + x2, and 00 yields information concerning only a filtration of q (x1; x2) = (x1 + x2)x2. For larger, even values the abelian groups which the spectral sequence is of n, further quadratic forms may be made by 0 00 said “to compute”. In algebraic topology, informa- applying one of q; q , or q to (x2i−1; x2i) for tion squeezed from invariants may be hard won. i = 1; 2; : : : ; n=2 and adding the results in F2. One’s motto should be “Do not expect too much Two quadratic forms q1 and q2 on V are called equivalent if there is a bijective, linear change of an invariant.” We shall return to the ASS later. Algebraic topology is generally believed to have of coordinates which transforms q1 into q2. For example, when n = 2, q00 is equivalent to q via the begun with Poincaré in [26], which initiated the study of differentiable manifolds. Poincaré posed coordinate transformation (x1; x2) , (x1 + x2; x2). A nonsingular quadratic form q1 is one for problems (e.g., the Poincaré Conjecture [24]) of which n is even and is equivalent either to the sum generalizing to higher dimensions the success of of n=2 copies of q or to q0 plus the sum of (n−2)/2 the nineteenth-century geometers in classifying copies of q. Hence, for each even integer n, there surfaces. His interest in manifolds stemmed partly are just two equivalence classes of q1’s. The Arf from his study of the global properties of solution invariant c(q1) lies in F2 and answers the question: curves to differential equations on orientable To which equivalence class does the nonsingular surfaces and partly from his use of the method quadratic form q1 belong? of Riemann surfaces in connection with complex The definition of c(q1) given in [5] involves function theory. a complicated algebraic formula, which at first A differentiable manifold is a set of points sight is not even well defined. In [7] Bill Browder in which each point lies in a coordinate patch used an amusing equivalent definition of the Arf modelled on the Euclidean space Rn consisting of invariant as the following “democratic invariant”. n-tuples of real numbers, which is familiar from The elements of V “vote” for either 0 or 1 by the several-variable calculus courses. Where two coor- function q1. The winner of the election (which is dinate patches overlap, the change-of-coordinates never a tie) is c(q1). Here is a table illustrating function is required to be highly differentiable 0 00 this for the three possibilities q; q ; q when in terms of the two sets of local coordinates. A V = f(0; 0), (0; 1), (1; 0), (1; 1)g. Having equal Arf differentiable map between two manifolds is a 00 invariants, q and q are equivalent, as we observed function f : M -! N that is differentiable in terms earlier, the vote being three to one in each case. of the local coordinates, and f is a diffeomorphism if it is one-one and onto and has a differentiable x (0; 0) (0; 1) (1; 0) (1; 1) value of c inverse map.
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